# Convergent sequence as series, maximum of sequence as limit

I'm currently studying for my math exams. I came across two exercises about sequences and series for which I have no clue. So any hints would be appreciated.

First problem:
$(a_n)_{n\in\mathbb{N}}$ is a convergent sequence. Show that:

$\lim\limits_{n\to \infty }\frac{1}{n}\sum\limits_{j=1}^na_j\:=\:\lim\limits_{n\to \infty }\left(a_n\right)$

Second:
Let $a_j \geq 0$ for $j = 1,\ldots,p$. Show that:
$\lim\limits_{n\to \infty }\left(a_1^n+...+a^n_p\right)^{\frac{1}{n}}\:=\:max\left\{a_1,\:...,\:a_p\right\}$

Hints:

a) Use the Stolz-Cesaro Theorem with $$\begin{cases}c_n:=a_1+\ldots+a_n\\b_n=n\end{cases}\;\;\implies\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=a_{n+1}\;\;\text{...and etc.}$$

(b) Use the squeeze theorem: WLOG, $\;a_1=\max\{a_1,..,a_p\}\;$ , so

$$a_1\le\sqrt[n]{a_1^n+\ldots+a_p^n}\le\sqrt[n]{pa_1^n}$$

First problem:

Let $a$ denote the limit of sequence $\left(a_{n}\right)$ and realize that the convergence of the sequence implies the existence of some $c>0$ such that $\left|a_{j}-a\right|<c$ for each $j$.

For $\epsilon>0$ find some $n_{0}$ such that $j>n_{0}\Rightarrow\left|a_{j}-a\right|<\frac{1}{2}\epsilon$.

For $n>n_{0}$ we have $\left|\frac{1}{n}\sum_{j=1}^{n}a_{j}-a\right|\leq\frac{1}{n}\sum_{j=1}^{n}\left|a_{j}-a\right|<\frac{1}{n}\left(n_{0}c+\frac{1}{2}\left(n-n_{0}\right)\epsilon\right)<\frac{n_{0}}{n}c+\frac{1}{2}\epsilon$.

Then find some $n_{1}\geq n_{0}$ s.t. $n>n_{1}\Rightarrow\frac{n_{0}}{n}c<\frac{1}{2}\epsilon$ so that: $$n>n_{1}\Rightarrow\left|\frac{1}{n}\sum_{j=1}^{n}a_{j}-a\right|<\epsilon$$